SOLUTIONS FOR THE TRAINING FINAL Remember : the final exam
... Remember : the final exam is cumulutative, though there will be more questions on the last third of the peogram. This is a set of training exercises on that last third (Groups action, rings and fields). 1.– Let R be a ring with unity, R∗ the groups of unit. a.– Show that the application R∗ × R → R, ...
... Remember : the final exam is cumulutative, though there will be more questions on the last third of the peogram. This is a set of training exercises on that last third (Groups action, rings and fields). 1.– Let R be a ring with unity, R∗ the groups of unit. a.– Show that the application R∗ × R → R, ...
Lecture 6 1 Some Properties of Finite Fields
... Claim 1 Let K and L be finite fields with K ⊆ L. Then, L can be viewed as a finite-dimensional vector space over K. Proof Idea Addition in the K-vector space is the addition law in L and scalar multiplication of an element α in L by an element c of K is defined to be the product cα as multiplied in ...
... Claim 1 Let K and L be finite fields with K ⊆ L. Then, L can be viewed as a finite-dimensional vector space over K. Proof Idea Addition in the K-vector space is the addition law in L and scalar multiplication of an element α in L by an element c of K is defined to be the product cα as multiplied in ...
![On the number of polynomials with coefficients in [n] Dorin Andrica](http://s1.studyres.com/store/data/023344871_1-db0421bb4ddb78289c0a283cc0aa29ca-300x300.png)
On the number of polynomials with coefficients in [n] Dorin Andrica
... Some interesting prime numbers Alexandru Gica University of Bucharest Abstract The aim of our research is to find all primes p which have the property that p+x2 has at most two prime divisors for all odd integers x such that x2
... Some interesting prime numbers Alexandru Gica University of Bucharest Abstract The aim of our research is to find all primes p which have the property that p+x2 has at most two prime divisors for all odd integers x such that x2
Math 113 Final Exam Solutions
... 2. a) (6 points) Let H = h(2, 4)i be a subgroup of Z × Z. Show that the cosets of H in Z × Z are precisely those of the form (0, n)H and (1, n)H, where n can be any integer. Suppose (r, n)H = (s, m)H where s, r = 0 or 1. Then we have (r − s, n − m) ∈ H. Note that |r − s| = 0 or 1. At the same time, ...
... 2. a) (6 points) Let H = h(2, 4)i be a subgroup of Z × Z. Show that the cosets of H in Z × Z are precisely those of the form (0, n)H and (1, n)H, where n can be any integer. Suppose (r, n)H = (s, m)H where s, r = 0 or 1. Then we have (r − s, n − m) ∈ H. Note that |r − s| = 0 or 1. At the same time, ...
PDF
... But Z8 is a bad ring, in the sense that non-zero elements can multiply to give 0 (2×4 = 0 here). As for another example, look at the set of all number ...
... But Z8 is a bad ring, in the sense that non-zero elements can multiply to give 0 (2×4 = 0 here). As for another example, look at the set of all number ...
Some proofs about finite fields, Frobenius, irreducibles
... K by grouping them in d-tuples of roots of elements of irreducible monic polynomials with coefficients in k = Fq , where d runs over positive divisors of n including 1 and n. Let Nd be the number of irreducible monic polynomials of degree d with coefficients in k = Fq . Then this grouping and counti ...
... K by grouping them in d-tuples of roots of elements of irreducible monic polynomials with coefficients in k = Fq , where d runs over positive divisors of n including 1 and n. Let Nd be the number of irreducible monic polynomials of degree d with coefficients in k = Fq . Then this grouping and counti ...
